Question

A rectangular field is to be enclosed on four sides with a fence. Fencing costs $2 per foot for two opposite sides, and $3 per foot for the other two sides. Find the dimensions of the field of area 830 ft2 that would be the cheapest to enclose.

Answer 1

To find the dimensions of the field that would be the cheapest to enclose, we need to minimize the total cost of the fencing. By analyzing the cost equation, we can determine that the minimum cost occurs when one pair of opposite sides has zero length. Substituting this value into the equation yields the dimensions of the field: 415 ft by 0 ft.

Step-by-step explanation:

To find the dimensions of the field that would be the cheapest to enclose, we need to minimize the total cost of the fencing. Let's assume the length of the field is 'x' and the width is 'y'. The total cost can be expressed as:

2x + 2y = A

where A is the area of the field, given as 830 ft2. Rearranging the equation, we have:

y = (A - 2x) / 2

Now, let's substitute the value of 'y' into the cost equation:

Cost = 2x + 3((A - 2x) / 2)

Simplifying further:

Cost = 2x + 3(A - 2x)/2

Cost = 2x + 3A/2 - 3x

Cost = -x + 3A/2

Now, we can plot a graph of the cost function and find the minimum point using calculus or we can apply logic. Since the coefficient of 'x' is negative, the cost will decrease as 'x' increases. So, the minimum cost will occur at 'x = 0', which means one pair of opposite sides has zero length. Substituting 'x = 0' into the equation for 'y', we can find the dimensions of the field:

y = (A - 2x) / 2

y = (830 - 2(0)) / 2

y = 415 ft

Therefore, the dimensions of the field that would be the cheapest to enclose are 415 ft by 0 ft.

Answer 2

Dimension for cheap enclose = 32.45 ft x 23.52 ft

Step-by-step explanation:

Area of rectangular field, A = 830 ft²

Length = l

Width = w

So we have

l x w = 830

`l=(830)/(w)`

Fencing costs $2 per foot for two opposite sides, and $3 per foot for the other two sides.

Cost for fencing, C = 2 x 2 x w + 3 x 2 x l = 4 w + 6 l

`C=4w+6* (830)/(w)`

For minimum cost we have derivative is zero

`dC=4* dw-6* (830)/(w^2)* dw\\\\0=4* dw-6* (830)/(w^2)* dw\\\\w^2=1245\\\\w=32.45ft\\\\lw=830\\\\l* 32.45=830\\\\l=23.52ft`

Dimension for cheap enclose = 32.45 ft x 23.52 ft